Answer:
Step-by-step explanation:
Given the limit of a function expressed as 
, to evaluate the following steps must be carried out.
Step 1: substitute x = 0 into the function
Step 2: Apply L'Hôpital's rule, by differentiating the numerator and denominator of the function
![= \lim_{ x\to \ 0} \dfrac{\frac{d}{dx}[ sin(x)-tan(x)]}{\frac{d}{dx} (x^3)}\\= \lim_{ x\to \ 0} \dfrac{cos(x)-sec^2(x)}{3x^2}\\](https://tex.z-dn.net/?f=%3D%20%5Clim_%7B%20x%5Cto%20%5C%200%7D%20%5Cdfrac%7B%5Cfrac%7Bd%7D%7Bdx%7D%5B%20sin%28x%29-tan%28x%29%5D%7D%7B%5Cfrac%7Bd%7D%7Bdx%7D%20%28x%5E3%29%7D%5C%5C%3D%20%5Clim_%7B%20x%5Cto%20%5C%200%7D%20%5Cdfrac%7Bcos%28x%29-sec%5E2%28x%29%7D%7B3x%5E2%7D%5C%5C)
Step 3: substitute x = 0 into the resulting function
Step 4: Apply L'Hôpital's rule, by differentiating the numerator and denominator of the resulting function in step 2
![= \lim_{ x\to \ 0} \dfrac{\frac{d}{dx}[ cos(x)-sec^2(x)]}{\frac{d}{dx} (3x^2)}\\= \lim_{ x\to \ 0} \dfrac{-sin(x)-2sec^2(x)tan(x)}{6x}\\](https://tex.z-dn.net/?f=%3D%20%5Clim_%7B%20x%5Cto%20%5C%200%7D%20%5Cdfrac%7B%5Cfrac%7Bd%7D%7Bdx%7D%5B%20cos%28x%29-sec%5E2%28x%29%5D%7D%7B%5Cfrac%7Bd%7D%7Bdx%7D%20%283x%5E2%29%7D%5C%5C%3D%20%5Clim_%7B%20x%5Cto%20%5C%200%7D%20%5Cdfrac%7B-sin%28x%29-2sec%5E2%28x%29tan%28x%29%7D%7B6x%7D%5C%5C)
Step 6: Apply L'Hôpital's rule, by differentiating the numerator and denominator of the resulting function in step 4
![= \lim_{ x\to \ 0} \dfrac{\frac{d}{dx}[ -sin(x)-2sec^2(x)tan(x)]}{\frac{d}{dx} (6x)}\\= \lim_{ x\to \ 0} \dfrac{[ -cos(x)-2(sec^2(x)sec^2(x)+2sec^2(x)tan(x)tan(x)]}{6}\\\\= \lim_{ x\to \ 0} \dfrac{[ -cos(x)-2(sec^4(x)+2sec^2(x)tan^2(x)]}{6}\\](https://tex.z-dn.net/?f=%3D%20%5Clim_%7B%20x%5Cto%20%5C%200%7D%20%5Cdfrac%7B%5Cfrac%7Bd%7D%7Bdx%7D%5B%20-sin%28x%29-2sec%5E2%28x%29tan%28x%29%5D%7D%7B%5Cfrac%7Bd%7D%7Bdx%7D%20%286x%29%7D%5C%5C%3D%20%5Clim_%7B%20x%5Cto%20%5C%200%7D%20%5Cdfrac%7B%5B%20-cos%28x%29-2%28sec%5E2%28x%29sec%5E2%28x%29%2B2sec%5E2%28x%29tan%28x%29tan%28x%29%5D%7D%7B6%7D%5C%5C%5C%5C%3D%20%5Clim_%7B%20x%5Cto%20%5C%200%7D%20%5Cdfrac%7B%5B%20-cos%28x%29-2%28sec%5E4%28x%29%2B2sec%5E2%28x%29tan%5E2%28x%29%5D%7D%7B6%7D%5C%5C)
Step 7: substitute x = 0 into the resulting function in step 6
![= \dfrac{[ -cos(0)-2(sec^4(0)+2sec^2(0)tan^2(0)]}{6}\\\\= \dfrac{-1-2(0)}{6} \\= \dfrac{-1}{6}](https://tex.z-dn.net/?f=%3D%20%20%5Cdfrac%7B%5B%20-cos%280%29-2%28sec%5E4%280%29%2B2sec%5E2%280%29tan%5E2%280%29%5D%7D%7B6%7D%5C%5C%5C%5C%3D%20%5Cdfrac%7B-1-2%280%29%7D%7B6%7D%20%5C%5C%3D%20%5Cdfrac%7B-1%7D%7B6%7D)
<em>Hence the limit of the function </em>
.
Answer:
Shira: x = 8
Samuel: m = 0
Step-by-step explanation:
Shira's mistake was that she subtracted 2 from both sides instead of adding to on both sides.
Correct Solving:
2x - 2 = 14
Add 2 to both sides;
2x = 16
Divide both sides by 2;
x = 8
Samuel's mistake was that when he distributed -2 to 8m and 8 he put the wrong sign for -2 * 8.
Correct Solving:
-2(8m + 8) = -16
Distribute;
-16m - 16 = -16
Add 16 to both sides;
-16m = 0
Divide both sides by -16;
m = 0
Hello :
13π/4 = (16-3)π/4
= 16π/4 -3π/4
13π/4 = -3π/4 +4π
tan( 13π/4) = tan ( -3π/4)= - tan ( 3π/4) =- tan( π - π/4) = - tan(- π/4)
tan( 13π/4) = -(-tan(π/4)) =tan(π/4)=1
Hello,
Answer is Dsince y=k(x-2)²-5 and if k=1 ==>D.
The slope = 17/8, while as the y - intercept = -5. The graph shown below will help give you a better understanding of what I'm saying.
